Question:
If \(2\sin^2 A + \sqrt{3}\cos A – 2 = 0\), then find the value of \(\tan A\).
यदि \(2\sin^2 A + \sqrt{3}\cos A – 2 = 0\) है, तो \(\tan A\) का मान ज्ञात कीजिए।
Options:
- \(\sqrt{3}\)
- \(1/\sqrt{3}\)
- 1
- 0
- \(\sqrt{3}\)
- \(1/\sqrt{3}\)
- 1
- 0
Correct Answer: b)
Explanation:
Replace \(\sin^2 A = 1 – \cos^2 A\):
\(2(1 – \cos^2 A) + \sqrt{3}\cos A – 2 = 0 \implies -2\cos^2 A + \sqrt{3}\cos A = 0\).
\(\cos A(\sqrt{3} – 2\cos A) = 0\).
For acute angle, \(\cos A = \dfrac{\sqrt{3}}{2} \implies A = 30^\circ\).
\(\tan A = \tan 30^\circ = \dfrac{1}{\sqrt{3}}\).
\(2(1 – \cos^2 A) + \sqrt{3}\cos A – 2 = 0 \implies \cos A(\sqrt{3} – 2\cos A) = 0\)।
\(\cos A = \dfrac{\sqrt{3}}{2} \implies A = 30^\circ\)।
\(\tan A = \tan 30^\circ = \dfrac{1}{\sqrt{3}}\)।